Question

sin^2x-10sinxcosx+21cos^2x=0

Answer 1

$sin^2(x)-10sin(x)cos(x)+21cos^2(x)=0\\x\neq \frac{\pi}{2}+2\pi k\\tg^2(x)-10tg(x)+21=0\\D=25-21=4\\tg(x)=3=>x=arctg(3)+\pi k\\tg(x)=7=>x=arctg(7)+\pi k$

k∈Z